Question

Answer 1

\(CuO+H_2\underrightarrow{t^o}Cu+H_2O\) 

0,1<----0,1<-- 0,1

\(Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)

0,05 <----0,15<-- 0,1

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

0,1 <------------------------- 0,1

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo pthh (hay bt Fe) có \(m_{Cu}=12-m_{Fe}=12-56.0,1=6,4\left(g\right)\Rightarrow n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)

\(\%_{m_{Fe_2O_3}}=\dfrac{160.0,05.100}{0,05.160+0,1.80}=50\%\)

\(\%_{m_{CuO}}=100\%-50\%=50\%\)

Theo PTHH: \(n_{H_2}=0,1+0,15=0,25\left(mol\right)\Rightarrow V_{H_2}=0,25.22,4=5,6\left(l\right)\)